已知sinA+sinB=sinC,cosA+cosB=cosC,求cos(A-B)的值

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sinA+sinB=sinC,cosA+cosB=cosC
(sinA)^2+(sinB)^2+2sinAsinB=(sinC)^2
(cosA)^2+(cosB)^2+2cosAcosB=(cosC)^2
(sinA)^2+(cosA)^2+(sinB)^2+(cosB)^2+2sinAsinB+2cosAcosB=(sinC)^2+(cosC)^2
1+1+2sinAsinB+2cosAcosB=1
sinAsinB+cosAcosB=-1/2
cos(A-B)=-1/2

{sinA+sinB=sinC (1)
{cosA+cosB=cosC (2)
(1)²+(2)²得
(sinA+sinB)²+(cosA+cosB)²=sin²C+cos²C
sin²A+2sinAsinB+sin²B+cos²A+2cosAcosB+cos&sup...

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{sinA+sinB=sinC (1)
{cosA+cosB=cosC (2)
(1)²+(2)²得
(sinA+sinB)²+(cosA+cosB)²=sin²C+cos²C
sin²A+2sinAsinB+sin²B+cos²A+2cosAcosB+cos²B=1
(sin²A+sin²B)+(cos²A++cos²B)+2(cosAcosB+sinAsinB)=1
1+1+2(cosAcosB+sinAsinB)=1
cosAcosB+sinAsinB=-1/2
cos(A-B)=-1/2

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